1.6 Resultant of coplanar non-concurrent force systems
After this lesson you should be able to:
- Reduce any coplanar force system to a resultant force, a resultant couple, or a force-couple pair at a chosen point.
- Locate the line of action of the resultant using Varignon's theorem.
- Replace distributed loads by equivalent concentrated resultants.
Theory
In most structures the forces do not meet at one point: loads sit at different positions along a beam, wind acts on a wall while self-weight acts through its centre, and reactions act at the supports. Such a system is non-concurrent. It can still be replaced by a simpler equivalent system with the same external effect on a rigid body.
The reduction has two steps. First, move every force to a chosen reference point O, adding the couple each force produces about O (lesson 1.5). Second, add the forces as vectors to get R and add the couples as signed scalars to get M_O. The original system is now equivalent to R acting at O plus the couple M_O.
There are three possible outcomes. If R is not zero, the force-couple pair can always be reduced further to R alone, shifted to a line of action at perpendicular distance d = |M_O| / R from O, on the side that reproduces the sense of M_O. If R is zero but M_O is not, the system is a pure couple. If both are zero, the system is in equilibrium. A practical way to locate the resultant is to find where its line of action crosses a convenient axis: for a crossing point on the x-axis, M_O = x Ry.
Distributed loads are treated the same way. A load of intensity w(x) along a beam has resultant equal to the area under the load diagram, acting through the centroid of that area. A uniform load w over length L gives wL at L/2; a triangular load rising from zero to w over L gives wL/2 at 2L/3 from the zero end. These substitutions are valid for finding reactions, but not for internal forces within the loaded length.
Derivation
- For forces F₁ … Fn at points (xi, yi): R = (ΣFxi, ΣFyi).
- Moment of the system about O: M_O = Σ (xi Fyi − yi Fxi).
- A single force R whose line of action crosses the x-axis at x̄ has moment about O equal to x̄ Ry (its x-component passes through the axis point at y = 0 and contributes y Rx = 0).
- Equating moments (Varignon): x̄ Ry = M_O, so x̄ = M_O / Ry. Similarly, the crossing of the y-axis is ȳ = −M_O / Rx.
- For a distributed load: R = ∫ w(x) dx (area) and x̄ = ∫ x w(x) dx / R (centroid).
R = √((ΣFx)² + (ΣFy)²); direction atan2(ΣFy, ΣFx)M_O = Σ (x Fy − y Fx)Crossing of x-axis: x̄ = M_O / Ry; perpendicular distance d = |M_O| / RUniform load: W = w L at L/2; triangular: W = w L / 2 at 2L/3 from the zero end
- Rigid body: the equivalent system reproduces external effects (reactions, overall motion) only.
- All forces in one plane.
- Distributed loads replaced by resultants only for whole-body equilibrium, not for internal forces within the loaded segment.
Codes and standards
Indian Standards are cited by number and clause only; no clause text, tables or figures are reproduced. Other standards are listed for comparison, not ranked.
- No design code governs this lesson. Load intensities used in examples (kN/m, kN/m²) would in practice come from IS 875 (Part 1):1987 and (Part 2):1987; EN 1991-1-1 and ASCE 7 serve the same role elsewhere. These are listed for comparison, not ranked.
Worked examples
Problem. A beam along the x-axis carries downward loads of 10 kN at x = 1 m, 20 kN at x = 3 m and 15 kN at x = 6 m. Find the resultant and its position.
- R = 10 + 20 + 15 = 45 kN downward.
- Moment about the left end (clockwise loads, taken as positive magnitudes here): 10 × 1 + 20 × 3 + 15 × 6 = 10 + 60 + 90 = 160 kN·m.
- x̄ = 160 / 45 = 3.556 m from the left end.
Answer. 45 kN downward at 3.556 m from the left end
Problem. A retaining structure is idealised with three forces in the x-y plane (x horizontal, y vertical, origin O at the toe): a 5 kN horizontal force (+x) acting at (0, 2 m), an 8 kN downward force at (3 m, 0) and a 4 kN upward force at (6 m, 0). Reduce the system to a single resultant and find where it crosses the x-axis.
- Rx = 5 kN; Ry = −8 + 4 = −4 kN.
- R = √(5² + 4²) = √41 = 6.403 kN, acting 38.66° below the +x direction.
- M_O = (0 × 0 − 2 × 5) + (3 × (−8) − 0) + (6 × 4 − 0) = −10 − 24 + 24 = −10 kN·m (clockwise).
- Crossing of the x-axis: x̄ = M_O / Ry = (−10) / (−4) = 2.5 m.
- Perpendicular distance from O to the line of action: d = 10 / 6.403 = 1.562 m.
Answer. R = 6.403 kN at 38.66° below horizontal, crossing the x-axis at 2.5 m from O
Step-by-step procedure
- Choose a reference point O, usually a support or a corner.
- Replace distributed loads by resultants at their centroids.
- Tabulate each force: Fx, Fy, coordinates and moment x Fy − y Fx.
- Sum to get Rx, Ry and M_O.
- Classify the outcome (single force, couple, or equilibrium) and locate the resultant with x̄ = M_O / Ry.
Site notes
- The resultant of soil and water pressure on a retaining wall, combined with the wall weight, must fall within the middle third of the base to avoid tension under the footing. This is the reduction of a non-concurrent system in daily use.
- Stacked materials on a slab are uneven; their resultant position tells you whether a beam or a cantilever edge is overloaded.
- For a crane on outriggers, the resultant of crane weight, counterweight and load must stay inside the outrigger footprint.
- Placing the resultant at the midpoint of the loads instead of using moments.
- Forgetting the moment of horizontal forces when taking moments about a point at ground level.
- Using wL/2 at L/2 for a triangular load; the resultant acts at the centroid, L/3 from the larger end.
- Using an equivalent concentrated load to compute internal shear or moment within the loaded length.
- Losing the sign of M_O and placing the resultant on the wrong side of O.
For parallel loads, the resultant always lies between the outermost loads if all act in the same direction. If your x̄ falls outside them, recheck the moment sum. With opposite-sense loads the resultant can lie outside, or the system may be a couple.
Summary
- Any coplanar system reduces to R at O plus M_O.
- If R ≠ 0, it reduces to a single force with x̄ = M_O / Ry.
- If R = 0 and M_O ≠ 0, it is a couple.
- Distributed load: resultant = area of load diagram, at its centroid.
Practice questions
Tap a question to see the answer.
1Downward loads of 30 kN at x = 0 and 50 kN at x = 4 m act on a beam. Locate their resultant.
R = 80 kN; x̄ = (50 × 4) / 80 = 2.5 m.
2A uniform load of 5 kN/m acts over 6 m. Give its resultant and position.
30 kN at 3 m from either end.
3A triangular load rises from zero at the left end to 6 kN/m at x = 3 m. Find its resultant and position.
W = 6 × 3 / 2 = 9 kN at 2 m from the left end.
4A 10 kN downward force at x = 0 and a 10 kN upward force at x = 2 m act on a beam. Reduce the system.
R = 0; M_O = 2 × 10 = +20 kN·m. It is a pure anticlockwise couple of 20 kN·m.
5Downward loads of 10 kN at x = 0 and 15 kN at x = 2 m and an upward load of 20 kN at x = 5 m act on a beam. Find the resultant and where it acts.
Ry = −5 kN; M_O = 2 × (−15) + 5 × 20 = +70 kN·m; x̄ = 70 / (−5) = −14 m, i.e. 5 kN downward acting 14 m to the left of the origin.
References
Standard texts, listed in no order of merit. Our explanations, examples and diagrams are original; nothing is copied from these books.
- F. P. Beer, E. R. Johnston, D. F. Mazurek, P. J. Cornwell, B. P. Self, Vector Mechanics for Engineers: Statics and Dynamics, McGraw Hill, 12th edition, 2019.
- R. C. Hibbeler, Engineering Mechanics: Statics, Pearson, 15th edition, 2022 (published 2021).
- J. L. Meriam, L. G. Kraige, J. N. Bolton, Engineering Mechanics: Statics, Wiley, 9th edition, 2018.
- S. Timoshenko, D. H. Young, J. V. Rao, S. Pati, Engineering Mechanics (in SI units), McGraw Hill Education (India), 5th edition, 2013.